Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

by
Knight, Randall D.

Published by
Pearson

ISBN 10:
0321740904

ISBN 13:
978-0-32174-090-8

Chapter 11 - Work - Exercises and Problems: 29

Answer

(a) $W = 9.8\times 10^5~J$
(b) $P = 1.96\times 10^4~watts$

Work Step by Step

(a) Since the elevator moves at a constant speed, the tension in the elevator cable is equal in magnitude to the weight of the elevator. We can find the work done by the motor.
$W = T~d$
$W = (mg)~d$
$W = (1000~kg)(9.80~m/s^2)(100~m)$
$W = 9.8\times 10^5~J$
(b) We can find the power supplied by the motor.
$P = \frac{W}{t}$
$P = \frac{9.8\times 10^5~J}{50~s}$
$P = 1.96\times 10^4~watts$